Problem: You have found the following ages (in years) of $4$ bears. Those bears were randomly selected from the $22$ bears at your local zoo: $ 5,\enspace 4,\enspace 6,\enspace 39$ Based on your sample, what is the average age of the bears? What is the standard deviation? Round your answers to the nearest tenth. Average age: $ $
Explanation: Because we only have data for a small sample of the $22$ bears, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$. To find the sample mean, add up the values of all $4$ samples and divide by $4$. $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{4}} x_i}{{4}} $ $ {\overline{x}} = \dfrac{5 + 4 + 6 + 39}{{4}} = {13.5\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated $({\overline{x}} = {13.5\text{ years}})$. Age $x_i$ Distance from the mean $(x_i - {\overline{x}})$ $(x_i - {\overline{x}})^2$ $5$ years $-8.5$ years $72.25$ years $^2$ $4$ years $-9.5$ years $90.25$ years $^2$ $6$ years $-7.5$ years $56.25$ years $^2$ $39$ years $25.5$ years $650.25$ years $^2$ Normally we can find the variance $({s^2})$ by averaging the squared deviations from the mean. But remember we don't know the real population mean—we had to estimate it by using the sample mean. The age of any particular bear in our sample is likely to be closer to the average age of the $4$ bears we sampled. This is compared to the average age of all $22$ bears in the zoo. Because of that, the squared deviations from the mean we calculated will probably underestimate the actual deviations from the population mean. To compensate for this underestimation, rather than simply averaging the squared deviations from the mean, we total them and divide by $n - 1$. $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{72.25} + {90.25} + {56.25} + {650.25}} {{4 - 1}} $ $ {s^2} = \dfrac{{869}}{{3}} = {289.67\text{ years}^2} $ The sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$. ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{289.67\text{ years}^2}} = {17\text{ years}} $ We can estimate that the average bear at the zoo is $13.5$ years old. There is also a standard deviation of $17$ years.